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Case I.  Suppose [-) to be positive.   Then U^  Ui ii
\ dx-Ii
positive and consequently Ui is a minimum. But according to the principle of the degradation of energy a body, which is free, moves in such a way as to diminish its potential energy Therefore when the force which produced the displacement 8c is removed the particle returns to the point (xi, yi, Zi), wher< its potential energy is a minimum. Evidently the equilib rium is stable in this case.
(d~U\ -1  to be negative.    Then U^  U dX"/i
is negative and consequently Ui is a maximum. Therefore if the particle is given a small displacement dx and then lef to itself, it will move away from the point (xi, yiy 21), wher< its potential energy is a maximum. In this case the equi librium is unstable.
Case III.  Suppose)-) to be zero.    There are thre-\to-/i
special cases to be considered:
(a)  The order of the first differential coefficient whicl does not vanish is odd.
(b)  The order of the first differential coefficient whicl does not vanish is even.
(c)  All of the differential coefficients vanish.
It is evident that when (c) is true the potential energy c the particle has a constant value and does not change witi the position of the particle. Therefore when the particl is left to itself after giving it a small displacement it wi neither return to its original position nor go on changin its position. The potential energy is the same and th particle is in equilibrium at all points. In this case th equilibrium is said to be neutral or indifferent.
It may be shown that when (a) holds the equilibrium i stable. On the other hand when (b) is true the equilibriur is stable or unstable according as the first differential coeffi cient which does not vanish is positive or negative.